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Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a temperature of 298 K. Assume that nitrogen comprises 78% of air by volume and that oxygen comprises 21%. Express your answers using two significant figures separated by a comma.

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Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Step-by-step explanation:

Ideal Gas Law → P . V = n . R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂ → Mole fraction of O₂ = 0.21

1% another gases → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles

We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

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